Mathematics

Evaluate the following integral
$$\int { \cfrac { \sin { 2x }  }{ a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x }  }  } dx$$


SOLUTION
$$\displaystyle\int{\dfrac{\sin{2x} \, dx}{a{\cos}^{2}{x}+b{\sin}^{2}{x}}}$$

Let
 $$t=a{\cos}^{2}{x}+b{\sin}^{2}{x}\Rightarrow\,dt=\left(-2a\cos{x}\sin{x}+2b\sin{x}\cos{x}\right)dx$$

$$\Rightarrow\,\dfrac{dt}{b-a}=2\sin{x}\cos{x}dx$$

$$\Rightarrow\,\dfrac{dt}{b-a}=\sin{2x}dx$$

$$\displaystyle\int{\dfrac{\sin{2x}dx}{a{\cos}^{2}{x}+b{\sin}^{2}{x}}}$$

$$=\dfrac{1}{b-a}\displaystyle\int{\dfrac{dt}{t}}$$

$$=\dfrac{1}{b-a}\log{\left|t\right|}+c$$

$$=\dfrac{1}{b-a}\log{\left|a{\cos}^{2}{x}+b{\sin}^{2}{x}\right|}+c$$ where $$t=a{\cos}^{2}{x}+b{\sin}^{2}{x}$$

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Subjective Medium Published on 17th 09, 2020
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